I am looking for a general method to find the area of convergence and sum function of the following complex series: $$ \sum_{n \geq 0} \frac{2^n z^{1+n}}{(2n)!} + \sum_{n<0} \frac{(-1)^n z^{3n}}{(n-1)}. $$ and $$ \sum_{n=1}^\infty n^2 z^{3n} + \sum_{n=0}^\infty \frac{z^{-n}}{(2n)!}. $$ For the first one, my attempt so far: $$ \sum_{n \geq 0} \frac{2^n z^{1+n}}{(2n)!} = z \sum_{n \geq 0} \frac{(2z)^n}{(2n)!} = \sum_{n \geq 0} \frac{(\sqrt{2z})^{2n}}{(2n)!} = z \cosh(\sqrt{2}\sqrt{z}), $$ which converges everywhere on $\mathbb{C}$, since $\cosh(x) = \frac{e^x + e^{-x}}{2}$. I'm also trying to use the Cauchy product on some known sequences like $e^z, \sin z, \cos z$, etc, but it doesn't seem to work so far, and it seems rather ad hoc and guessing. Is there a more general method that can be used to solve these?


The second sum change the sign of $n$, it is almost the sum for $\ln(1 + x^3)$. The third you'll get from the geometric series by taking derivative, then multiplying by $x$, twice (one round gives a factor $n$), the fourth you get from $\cos$.

No, there isn't an easy way to do this. But Wilf's generationfunctionology fools around with lots of (formal) series, and gives quite a cataloge of useful operations.

  • $\begingroup$ Making it positive gives $\sum_{n > 0} \frac{(-1)^{n+1} z^{-3n}}{n+1}$, is this correct? What is the sum for $\ln x^3$? I have not seen that as a sum series. $\endgroup$ – Sigurd Mar 4 '20 at 20:49
  • $\begingroup$ @Sigurd, for $\ln(1 + x^3)$, replace $x$ by $x^3$ $\endgroup$ – vonbrand Mar 5 '20 at 21:25

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